Calculate the ratio of catalyzed and uncatalyzed rate constant at 20 degree Celsius if the energy of activation of a catalyst reaction is 20 kilo joule per mole 4 and for an uncatalyst reaction is 75 kilo joule per mole?

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Answer 1 · 2017-09-20T16:52:54

#k_(cat)/k = 6.31 xx 10^9#

We are asked to calculate #k_(cat)/(k)# for a reaction at a fixed temperature. We can start from the Arrhenius equation:

#k = Ae^(-E_a//RT)#

Having a fixed #T# and two different activation energies #E_a# gives rise to two different rate constants #k_i#:

#k_(cat) = Ae^(-E_(a,cat)//RT)#

#k = Ae^(-E_a//RT)#

Assuming the frequency of collisions is the same in the different mechanism:

#k_(cat)/k = e^(-E_(a,cat)//RT)/e^(-E_a//RT)#

#= e^(-E_(a,cat)/(RT) + E_a/(RT))#

#= e^(-(E_(a,cat) - E_a)/(RT))#

As usual, we must have temperature in #"K"#...

#20^@ "C" = "293.15 K"#

Thus:

#color(blue)(k_(cat)/k) = e^(-("20 kJ/mol" - "75 kJ/mol")/("0.008314472 kJ/mol"cdot"K"cdot"293.15 K"))#

#= color(blue)(6.31 xx 10^9)#

This means your reaction is over a billion times faster due to the catalysis.

Why must we have #R# in #"kJ/mol"cdot"K"#?